How many real values of x are there for which (x - 1)2 + (x - 2)2 + (x - 3)2 + (x - 4)2 = 0
Explanation:
We know square of any real number is always greater than or equal to zero.
We have (x - 1)2 + (x - 2)2 + (x - 3)2 + (x - 4)2
Each of these terms is a square hence greater than or equal to zero.
Sum of these numbers will be equal to zero only when all of them are equal to zero.
But,
(x - 1) = 0 when x = 1 (x - 2) = 0 when x = 2 (x - 3) = 0 when x = 3 (x - 4) = 0 when x = 4
Hence, there is no value of x for which all of them are equal to zero at the same time.
∴ There is no value of x for which (x - 1)2 + (x - 2)2 + (x - 3)2 + (x - 4)2 = 0
Alternately, Given, (x - 1)2 + (x - 2)2 + (x - 3)2 + (x - 4)2 = 0
⇒ (x2 - 2x + 1) + (x2 - 4x + 4) + (x2 - 6x + 9) + (x2 - 8x + 16) = 0
⇒ 4x2 - 20x + 30 = 0
⇒ 2x2 - 10x + 15 = 0
D = (-10)2 - 4 × 2 × 15 = -20 < 0
Hence, there is no real root for (x - 1)2 + (x - 2)2 + (x - 3)2 + (x - 4)2 = 0
Hence, 0.
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